Slow and steady wins the race
A HARE AND A TORTOISE RACE ON A CIRCULAR TRACK OF RADIUS 50m. THEY BOTH START FROM THE SAME POINT BUT RUN IN OPPOSITE DIRECTIONS (CLOCKWISE AND COUNTER CLOCKWISE). THE FIRST ONE TO COMPLETE A FULL CIRCLE AND RETURN TO THE STARTING POINT WINS. ALSO, IT IS GIVEN THAT:
1)THE HARE STARTS RUNNING WHEN THE TORTOISE HAS ALREADY COVERED 1/5th OF THE DISTANCE.
2)THE HARE MEETS THE TORTOISE ON THE TRACK WHEN THE HARE HAS RUN 1/8th OF THE DISTANCE IN HIS DIRECTION.
BY WHAT FACTOR SHOULD THE HARE INCREASE ITS SPEED AFTER MEETING THE TORTOISE IF IT HOPES TO END THE RACE IN A TIE???
The answer is 37.8.
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1 solution
let the speed of the hare and the tortoise be v1 & v2 respectively
its given that they both meet on the track when the hare has run 1/8th of the distance or alternatively, the tortoise has run 7/8th of the distance
but its also given that the hare had started running when the tortoise had already run 1/5th of the distance...
therefore, time taken by hare to complete 1/8th distance=time taken by tortoise to cover (7/8th - 1/5th) of the distance
but distance for each contestant=2 pi 50=314m & time=distance/speed
therefore, [(1/8) 314]/v1 = [(27/40) 314]/v2 simplifying, we get v2=5.4*v1
let x be the factor by which hare accelerates after meeting the tortoise to tie the match now, the time taken by hare to complete the rest, i.e, 7/8th distance=time taken by tortoise to run 1/8th distance [(7/8) 314]/(x v1)=[(1/8) 314]/(5.4 v1)
solving the above equation, we get x=37.8